The answer is $\frac{1}{20^{2012}}$.
Let us treat more general case when $P$ is the set of all $n$ tuples. If $A = B = P$ then
$f(A, B) = \frac{1}{20^n}$. We prove that $f(A, B) \le \frac{1}{20^n}$ by induction over $n$.

$n = 1$. Suppose that $A = \{1, 2, \dots, a+c\}$, $B = \{20-b-c+1, \dots, 20\}$. Then $|A \cap B| = c$, $|A| = a+c$, $|B| = b+c$ and $f(A, B) = \frac{c}{(a+c)(b+c)} = \frac{1}{20+ab/c} \le \frac{1}{20}$.
Suppose that the statement is correct for $n-1$. Let $A = \bigcup_{i=1}^{20} A_i$, where elements of the set $A_i$ are obtained from elements of $A$ having last entry $i$ by removing this last entry. By definitions $|A| = \sum_{i=1}^{20} |A_i|$ and $A_1 \subset A_2 \subset \dots \subset A_{20}$. Let $B = \bigcup_{i=1}^{20} B_i$, where elements of the set $B_i$ are obtained from elements of $B$ having last entry $i$ by removing this last entry. By definitions $|B| = \sum_{i=1}^{20} |B_i|$ and $B_1 \supset B_2 \supset \dots \supset B_{20}$. Now
$$|A \cap B| = \sum_{i=1}^{20} |A_i \cap B_i| \le \frac{1}{20^{n-1}} \sum_{i=1}^{20} |A_i| \cdot |B_i| \le \frac{1}{20^{n-1}} \cdot \frac{1}{20} \left( \sum_{i=1}^{20} |A_i| \right) \left( \sum_{i=1}^{20} |B_i| \right) = \frac{1}{20^n} |A| \cdot |B|$$
(The first inequality is valid due to inductive hypothesis, the second inequality is the Chebyshev's rearrangement inequality). Done.